Given the area, find the dimension of a rectangle The width of a rectangle is 30 feet less than twice its length. its area is 68 square feet. Find the dimensions of the rectangle.

Length=17

Width=4

Step-by-step explanation:

Length=x

Width=2x-30

x(2x-30)=68

2x²-30x=68

2x²-30x-68=0

2(x²-15x-34)=0

2(x+2)(x-17)=0

So x equals either 17 or -2.  Since it can’t be -2, x=17

Length=x=17

Width=2x-30=2*17-30=34-30=4

Length  L = 17 ft.

Width   W = 4 ft

Step-by-step explanation:

Area of a rectangle = L * W

Area = 68 ft²

W = 2L - 30

Area = L * W

68 = L * (2L - 30)

68 = 2L² - 30L

2L² - 30L -68 = 0     ( solve L by using quadratic equation)

        -(-30) ± √ -30² - 4 * 2 * (-68)

L =

                       2 * 2

L = 17 ;  L= - 2   therefore use L = 17 ft.

solve for W:

W = 2L - 30

W = 2(17) - 30

W = 34 - 30

W = 4 ft

check:

Area = L * W

68 = 17 * 4

68 = 68   OK

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step-by-step explanation:

answer: the answer is one

step-by-step explanation:

pemdas, in this situation there is no parenthesis or exponents so you go to multiplication, 1x1=1, then division 1/1= 1 then addition 1+1=2 then subtraction 2-1=1